10.1 composition of functions functions modeling change: a preparation for calculus, 4th edition,...

10.1

COMPOSITION OF FUNCTIONS

Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Composition of Functions

• The function f(g(t)) is said to be a composition of f with g.

• The function f(g(t)) is defined by using the output of the function g as the input to f.

• The function f(g(t)) is only defined for values in the domain of g whose g(t) values are in the domain of f.


Formulas for Composite FunctionsExample 1

Let p(x) = sin x + 1 and q(x) = x2 − 3. Find a formula in terms of x for w(x) = p(p(q(x))).

Solution

We work from inside the parentheses outward. First we find p(q(x)), and then input the result to p.

w(x) = p(p(q(x)))

= p(p(x2 − 3))

= p(sin(x2 − 3)) + 1

= sin(sin(x2 − 3) + 1) + 1.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Composition of Functions Defined by Graphs

Example 3Let u and v be two functions defined by the graphs. Evaluate:

(a) v(u(−1)) (b) u(v(5)) (c) v(u(0)) + u(v(4))

Solution(a) To evaluate v(u(−1)), start with u(−1). From the figure, we see

that u(−1) = 1. Thus, v(u(−1)) = v(1). From the graph we see that v(1) = 2, so v(u(−1)) = 2.

(b) Since v(5) = −2, we have u(v(5)) = u(−2) = 0.(c) Since u(0) = 0, we have v(u(0)) = v(0) = 3. Since v(4) = −1, we

have u(v(4)) = u(−1) = 1. Thus v(u(0)) + u(v(4)) = 3 + 1 = 4.Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

3 2 1 1x

1

6

1 1 2 3 4 5 6x

3

1

1

3u(x) v(x)(-3,6)

(-1,1)

Composition of Functions Defined by TablesExample 2Complete the table. Assume that f(x) is invertible.SolutionWe will first look at g(f(2)). From the table we see that f(2) = 1. Therefore, we have g(f(2)) = g(1). Since g(1) = 1, we can fill in the entry for g(f(2)) the following way: g(f(2)) = g(1) = 1.To find g(2), we have to use information about g(f(2)). We first need to find a value of x such that f(x) = 2. That means that we are looking for f−1(2). From the table we see that f−1(2) = 0,or equivalently, f(0) = 2. Therefore, g(2) = g(f(0)). From the table we see that g(f(0) = 2. Thus, g(2) = 2.


x f(x) g(x) g(f(x))0 2 3 21 3 1 32 1

Decomposition of FunctionsExample 4Let h(x) = f(g(x)) = . Find possible formulas for f(x) and g(x).SolutionIn the formula h(x) = , the expression x2 + 1 is in the exponent. We can take the inside function to be g(x) = x2 + 1. This means that we can write

Then the outside function is f(x) = ex. We check that composing f and g gives h:

There are many possible solutions to Example 4. For example, we might choose f(x) = ex+1 and g(x) = x2. Then


12xe

12xe

)(12

)( xgx eexh

)()1())(( 12 2

xhexfxgf x

)())(( 11)( 2

xheexgf xxg

10.2

INVERTIBILITY AND PROPERTIES OF INVERSE

FUNCTIONS


Definition of Inverse Function

Suppose Q = f(t) is a function with the property that each value of Q determines exactly one value of t. Then f has an inverse function, f−1, and

f−1(Q) = t if and only if Q = f(t).

If a function has an inverse, it is said to be invertible.


Finding a Formula for an Inverse FunctionExample 3Find the inverse of the function f(x) = 3x/(2x + 1).SolutionFirst, we solve the equation y = f(x) for x:

y = 3x/(2x + 1) 2xy + y = 3x 2xy – 3x = – y x (2y – 3) = – y

x = – y/(2y – 3) x = y/(3 – 2y)

As before, we write x = f-1(y) = y/(3 – 2y).Since y is now the independent variable, by convention we rewrite the inverse function with x as the independent variable. We have y = f-1(x) = x/(3 – 2x).


Noninvertible Functions: Horizontal Line Test

The Horizontal Line Test

If there is a horizontal line which intersects a function’s graph in more than one point, then the function does not have an inverse. If every horizontal line intersects a function’s graph at most once, then the function has an inverse.


The graph of q(x) = x2 fails the horizontal line test, so q(x) = x2 has no inverse.

3 3x

9

y q(x) = x2

Graphing A Function And Its InverseExample 5Let P(x) = 2x.

(a) Show that P is invertible. (b) Find a formula for P−1(x).(c) Sketch the graphs of P and P−1 on the same axes.(d) What are the domain and range of P and P−1?

Solution(a) Since P is an exponential function with base 2, it is always increasing, and

therefore passes the horizontal line test. (b) To find a formula for P−1(x), we solve for x in the equation 2x = y. We take the

log of both sides and simplify to get x = (log y)/(log 2) or switching variables: y = P-1(x) = (log x)/(log 2)

(c) Tables of values for P(x) and P-1(x). Interchanging the rows of P(x) gives P-1(x).


x -3 -2 -1 0 1 2 3P(x) 0.125 0.25 0.5 1 2 4 8

x 0.125 0.25 0.5 1 2 4 8P-1(x) -3 -2 -1 0 1 2 3

Graphing A Function And Its InverseExample 5 continuedLet P(x) = 2x.

(c) Sketch the graphs of P and P−1 on the same axes.(d) What are the domain and range of P and P−1?

Solution(b) y = P-1(x) = (log x)/(log 2) (c) From the tables of values for P(x) and P-1(x), we have the

graphs on the right.Notice how they are mirrorimages of each other throughthe line y = x

(d) The domain of P, an exponential function, is all real numbers, and its range is all positive numbers. The domain of P−1, a logarithmic function, is all positive numbers and its range is all real numbers.


3 1 4 8x

3

1

4

8y

y = x

P(x) = 2xP-1(x) = (log x)/(log 2)

The Graph, Domain, and Range of an Inverse Function

• Graph of f−1 is reflection of graph of f across the line y = x.

• Domain of f−1 = Range of f• Range of f−1 = Domain of f• Example:


x

y

f

f−1

y = x

A Property of Inverse Functions

If y = f(x) is an invertible function and y = f−1(x) is its inverse, then

• f−1(f(x)) = x for all values of x for which f(x) is defined,

• f(f−1(x)) = x for all values of x for which f−1(x) is defined.


Example 6(a)(a) Check that f(x) = x/(2x + 1) and f−1(x) = x/(1 − 2x) are inverse

functions of each other.Solution (a)

Similarly, you can check that f (f−1(x)) = x.

x

x

xx

xx

xx

xx

xx

xx

xf

xfxff

12112

122

1212

12

1221

12)(21

)())((1

A Property of Inverse Functions


Example 6(b)f(x) = x/(2x + 1) and f−1(x) = x/(1 − 2x)(b) Graph f and f−1 on axes with the same scale. What are the

domains and ranges of f and f−1?Solution (b)

Properties of Inverse Functions


x

y

f−1(x) = x/(1 − 2x) with asymptotesy = -½ and x = ½

f(x) = x/(2x + 1)with asymptotesy = ½ and x = -½

The domain of f(x) is all real numbers except ½ and the range is all real numbers except - ½

The domain for f−1(x) is all real numbers except -½ and the range is all real numbers except ½

y =x

Restricting the Domain


A function that fails the horizontal line test is not invertible. For this reason, the function f(x) = x2 does not have an inverse function. However, by considering only part of the graph of f, we can eliminate the duplication of y-values. Suppose we consider the half of the parabola with x ≥ 0. This part of the graph does pass the horizontal line test because there is only one (positive) x-value for each y-value in the range of f.

x

y The graphs of f and f−1 are shown in the figure. Note that the domain of f is the the range of f−1, and the domain of f−1 (x ≥ 0) is the range of f.

f(x) = x2

xxf )(1

Inverse Trigonometric Functions

In Section 8.4 we restricted the domains of the sine, cosine, and tangent functions in order to define their inverse functions:

y = sin−1 x if and only if x = sin y and −π/2 ≤ y ≤ π/2

y = cos−1 x if and only if x = cos y and 0 ≤ y ≤ π

y = tan−1 x if and only if x = tan y and −π/2 < y < π/2 .


10.3

COMBINATIONS OF FUNCTIONS


The Difference of Two Functions Defined by Formulas: A Measure of Prosperity

Consider the population function P(t) = 2 (1.04)t and the number of people that a country can feed N(t) = 4 + 0.5 t where t is measured in years and both P and N represent millions of people.


50 100tyea rs

50

100

m i lli o n s o f p eop le

Shortages occur

(78.32,43.16)

47.226 80tyea rs2

14.865m i lli o n s o f p eop le

We explore this situation by plotting these two functions on the same graph and see when the population exceeds the number who can be fed.

We could also graph the difference N(t) – P(t) and observe when the maximum surplus occurs.

Point of maximum surplus

The Sum and Difference of Two Functions Defined by Graphs


Example 1Let f(x) = x and g(x) = 1/x. By adding vertical distances on the graphs of f and g, sketch h(x) = f(x) + g(x) for x > 0.Solution The graphs of f and g are shown in the figure. For each value of x, we add the vertical distances that represent f(x) and g(x) to get a point on the graph of h(x). Compare the graph of h(x) to thevalues shown in the table.x ¼ ½ 1 2 4f(x) = x ¼ ½ 1 2 4g(x) = 1/x 4 2 1 ½ ¼ h(x) = f(x) + g(x) 4 ¼ 2 ½ 2 2 ½ 4 ¼

0 1 2 3 4 5 6x

1

2

3

4

5

6y

Note that as x increases, g(x) decreases toward zero, so the values of h(x) get closer to the values of f(x). On the other hand, as x approaches zero, h(x) gets closer to g(x).

y = h(x)

y = g(x)

y = f(x)

Example 6Find exactly all the zeros of the function

p(x) = 2x · 6x2 − 2x · x − 2x+1.SolutionWe rewrite the formula for p as

p(x) = 2x · 6x2 − 2x · x − 2x+1

= 2x(6x2 − x − 2) factoring out 2x

= 2x(2x + 1)(3x − 2) factoring quadratic

Since p is a product, it equals zero if one or more of its factors equals zero. But 2x is never equal to 0, so p(x) equals zero if and only if one of the linear factors is zero:

(2x + 1) = 0 or (3x − 2) = 0So x = −1/2 or x = 2/3

Factoring a Function’s Formula into a Product


Tables for Example 3

The Quotient of Functions Defined by Tables: Per Capita Crime Rate


Year 2005 2006 2007 2008 2009 2010Years since 2005 0 1 2 3 4 5

Crimes in City A = NA(t) 793 795 807 818 825 831

Crimes in City B = NB(t) 448 500 525 566 593 652

Population of City A = PA(t) 61,000 62,100 63,220 64,350 65,510 66,690

Population of City B = PB(t) 28,000 28,588 29,188 29,801 30,427 31,066

rA(t) = NA(t) / PA(t) 0.013 0.0128 0.01276 0.01271 0.01259 0.01246rB(t) = NB(t) / PB(t) 0.016 0.01749 0.01799 0.01899 0.01949 0.02099

We see that between 2005 and 2010, City A has a lower per capita crime rate than City B. Furthermore, the crime rate of City A is decreasing, whereas the crime rate of City B is increasing. Thus, even though the table indicates that there are more crimes committed in City A, it appears that City B is, in some sense, more dangerous. The table also tells us that, even though the number of crimes is rising in both cities, City A is getting safer, while City B is getting more dangerous.

10.1 composition of functions functions modeling change: a preparation for calculus, 4th edition,...

Documents

function fx

function g

value of x

inside function

thenfunctions modeling

outside function

terms of x

xy y